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I get the idea behind partial differentiation but this one is really tricky!

If $z = xe^{-y}$, and $x = \cosh t$, and $y = \cos s$, then what is the partial of $z$ with respect to $s$, and partial of $z$ with respect to $t$?

Thanks

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2 Answers

There are two ways to do this:

1.) just plug in the expressions for $x$ and $y$ in terms of $s$ and $t$ and differentiate.

2.) use the multivariate chain rules:

$$ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} $$ $$ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t} $$

Make sense?

By request, $$ dz = d(xe^{-y}) =e^{-y}dx+-xe^{-y}dy$$ however, since $x = \cosh(t)$ and $y = \cos(s)$ we find $$ dx = \sinh(t)dt \qquad \text{and} \qquad dy = -\sin(s)ds $$ plug those into the $dz$ formula to obtain: $$ dz = e^{-y}\sinh(t)dt -xe^{-y}(-\sin(s)ds) $$ We can read from the above the coefficient of $dt$ is $\frac{\partial z}{\partial t}$ and the coefficient of $ds$ is $\frac{\partial z}{\partial s}$. Of course I use $x$ and $y$ here as abbreviations for the $t,s$ formulas. Intuitively, the partial derivative w.r.t. $t$ is when $s$ is held constant so $ds=0$ and this is why this approach works. In advanced calculus we can give better answers in terms of the implicit or inverse function theorems. As a general principle, you can use differentials and proceed formally, this approach goes a long way.

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If you want to know why these are the correct formulas I can elaborate. –  James S. Cook Sep 15 '12 at 2:28
    
It makes sense and partial of x w.r.t. s = 0 and partial of y w.r.t. t = 0 –  mary Sep 15 '12 at 21:32
    
@mary correct, I'm just giving the answer that works no matter what the formulas you get for x and y. You said "differentials" in the original post, did you want to see more about that? I known a way to do all of this formally by pushing differentials around formally.... if you're interested. –  James S. Cook Sep 16 '12 at 2:27
    
Please go ahead and show me. Thans –  mary Sep 17 '12 at 4:36
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@MykeArya not really, I understood the method of differentials far before I understood the exterior calculus. Moreover, I also gave your answer in my original post, granted you have included a few details which are nice. –  James S. Cook Apr 12 '13 at 17:42
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$\frac{\partial{z(x,y)}}{\partial{s}}=\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}+\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{s}}=\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}+0=\{-xe^{-y}\}\{-\sin(s)\}$

$\frac{\partial{z(x,y)}}{\partial{t}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{t}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{t}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{t}}+0=\{e^{-y}\}\{-\sinh(s)\}$

It's just applying chain rule and noticing that $\frac{\partial{x}}{\partial{s}}=0$ because $x$ does not depend on $s$. Therfore is constant regarding this variable, and the derivative of a constant vanishes. And similarty $\frac{\partial{y}}{\partial{t}}=0$.

The trick when doing this kind of things, is realizing that partial differenciation means that the other variables are constants regarding this operation. Otherwise is just being careful and not to miss anything in the steps.

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